Math Revisited: The Cauchy-Riemann Equations And Harmonic Functions

Time to revisit one of the more fundamental aspects of advanced calculus. Given a function defined:

f(z) = f(x + iy)   = u(x,y) + iv(x,y)

 

which is taken to be differentiable at the point z ₀ = x₀ + iy₀  - then the partial derivatives of u and v exist at the point (x₀ ,  y₀ ) and satisfy the equations:

 

a) ¶ u/ ¶ x  =  ¶ v/ ¶ y   and   b)  ¶ v/ ¶ x  =  -  ¶ u/ ¶ y  

Which are the Cauchy- Riemann equations. If such condition is met then the function is said to be analytic in the region, Â.  

An additional condition likely to be examined is whether the function is harmonic. If it is, then u and v also have continuous 2^nd partial derivatives. Then, we can differentiate both sides of (a) with respect to x, and (b) with respect to y to obtain:

 

1) ¶ ² u/ ¶ x²  =    ¶ ² v/ ¶ x  ¶ y    and

 

2)  - ¶ ² u/ ¶ y²  =    ¶ ² v/ ¶ y  ¶ x  

 

From which:

 ¶ ² u/ ¶ x²  =   - ¶ ² u/ ¶ y²       or    ¶ ² u/ ¶ x²  +  ¶ ² u/ ¶ y²       = 0

Which is known as Laplace’s equation.

Similarly, we can perform an analogous process for v (differentiating both sides of (a) with respect to y and (b) with respect to x) to arrive at:

 

¶ ² v/ ¶ x²  +  ¶ ² v/ ¶ y²       = 0

If then this equation is satisfied, v is harmonic.

 

Let’s look at the  function.

  f(z) = (x² – y²) + i2xy

So that: u(x,y) =  x² -  y²

 

And v(x,y) =  2x y

 

Now, we first check to see if the eqns. are analytic

 

Take ¶ u/ ¶ x    =   2x

 

And:  ¶ v/ ¶ y    =   2x

 

Since:  ¶ u/ ¶ x  =  ¶ v/ ¶ y     then u(x.y) is analytic

 

Now check the other function, v(x.y)

¶ v/ ¶ x  =  2y

 

And:  -  ¶ u/ ¶ y   =  - (-2y) = 2y

 

So that v(x,y) is analytic..

 

Note: If f(z) is analytic everywhere in the complex plane it is said to be an entire function.

 

Now, check to see whether the functions are harmonic.

 

For u(x,y) this means we need: ¶ ² u/ ¶ x²  +  ¶ ² u/ ¶ y²       = 0

Since: ¶ u/ ¶ x    =   2x, then   ¶ ² u/ ¶ x²  =   2

 

Since: ¶ u/ ¶ y   =   -2y then  ¶ ² u/ ¶ y²    =   -2 

Then:

¶ ² u/ ¶ x²  +  ¶ ² u/ ¶ y²       = 2 + (-2) = 0

 

For v(x.y): we need ¶ ² v/ ¶ x²  +  ¶ ² v/ ¶ y²       = 0

 

Since: ¶ v/ ¶ x  =  2y then  ¶ ² v/ ¶ x²  =   2

 

Since: ¶ v/ ¶ y    =   2x   then  ¶ ² v/ ¶ y²      = 2

Then: ¶ ² v/ ¶ x²  +  ¶ ² v/ ¶ y²         = 2 + 2 = 4

 

So, v(x,y) is evidently not harmonic since the sum of the 2^nd partials are not equal zero.

 

Problems for the Math Maven:

 

1)     Given the function: u(x,y) = x³ – 3xy²

 

Show the function is harmonic on the entire complex plane.

 

2)     Given the function: u(x.y) = exp(-x) [x sin y – y cos y]

 

a)     Show u(x,y) is harmonic

b)     Find v(x,y) such that f(z) = u + iv is analytic

 
3) Let f(z) = exp(x) cos(y) + i(exp(x)sin(y) = u(x,y) + iv(x,y)


a) Determine if the function is analytic for both u and v.

b) Determine if the function is harmonic for both u and v.